A double-stranded DNA fragment has a total of 10,000 base pairs. Analysis reveals that 20% of the nitrogenous bases are Adenine. Calculate the exact number of Thymine, Cytosine, and Guanine bases present in this DNA fragment. Also, determine the total length of this DNA molecule in nanometers, given that the distance between two consecutive base pairs is 0.34 nm.

Question

Accepted Answer

Step-by-Step Calculation

Step 1: Total number of nucleotides

Given total base pairs (bp) = 10,000
Since each base pair consists of 2 nucleotides, total number of nucleotides = $10,000 \times 2 = 20,000$.

Step 2: Calculate the number of Adenine (A) and Thymine (T) bases

According to Chargaff's rules, the percentage of Adenine equals Thymine ($A = T$), and Cytosine equals Guanine ($C = G$).
Given that Adenine = 20% of the total bases.
Therefore, Thymine (T) = 20% of the total bases.
Total percentage of (A + T) = $20% + 20% = 40%$.
Number of Adenine bases = $20% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = 4,000$.
Number of Thymine bases = $20% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = 4,000$.

Step 3: Calculate the number of Cytosine (C) and Guanine (G) bases

Total percentage of all bases = 100%
Percentage of (C + G) = $100% - 40% = 60%$.
Since $C = G$, the percentage of Cytosine = $30%$ and Guanine = $30%$.
Number of Cytosine bases = $30% \text{ of } 20,000 = \frac{30}{100} \times 20,000 = 6,000$.
Number of Guanine bases = $30% \text{ of } 20,000 = \frac{30}{100} \times 20,000 = 6,000$.

Step 4: Calculate the total length of the DNA molecule

Total base pairs = 10,000
Distance between two adjacent base pairs = $0.34 \text{ nm} = 0.34 \times 10^{-9} \text{ m}$.
Total length = $\text{Total base pairs} \times \text{Distance between two base pairs}$
Total length = $10,000 \times 0.34 \text{ nm} = 3,400 \text{ nm}$ (or $3.4 \mu\text{m}$).